Homotopy Type Theory
isomorphism in a precategory (Rev #6)

Definition

A morphism f:homA(a,b)f:hom_A(a,b) of a precategoryAA is an isomorphism if there is a morphism g:homA(b,a)g:hom_A(b,a) such that gf=1agf=1_a and fg=1bfg=1_b. We write a≅ba \cong b for the type of such isomorphisms.

If f:a≅bf:a \cong b, then we write f−1f^{-1} for its inverse which is uniquely determined by Lemma 9.1.3.

Properties

Lemma 9.1.3

For any f:homA(a,b)f : hom_A(a,b) the type “f is an isomorphism” is a mere proposition?. Therefore, for any a,b:Aa,b:A the type a≅ba \cong b is a set. Proof. Suppose given g:homA(b,a)g:hom_A(b,a) and η:1a=gf\eta:1_a = g f and ϵ:fg=1b\epsilon : f g=1_b, and similarly g′,η′g',\eta', and ϵ′\epsilon'. We must show (g,η,ϵ)=(g′,η′,ϵ′)(g,\eta,\epsilon)=(g',\eta', \epsilon'). But since all hom-sets are sets , their identity types are mere propositions, so it suffices to show g=g′g=g'. For this we have g′=1ag′=(gf)g′=g(fg′)=g1b=g□ g' = 1_{a} g'= (g f) g' = g (f g') = g 1_{b}= g \square